這是我的問題的人為示例，因此請忽略這是通過使用帶有 json 可選參數(shù)的奇異結(jié)構(gòu)來解決的。鑒于： { "name": "alice", "address_dict": { "home": { "address": "123 st" }, "work": { "address": "456 rd", "suite": "123"} } }type AddressIf interface{}type AddressHome { Address string `json:"address"`}type AddressWork { Address string `json:"address"` Suite string `json:"suite"`}type Contact struct { Name string `json:"name"` AddressMap map[string]AddressIf `json:"address_map"`}func(self *Contact) UnmarshalJSON(b []byte) error { var objMap map[string]*json.RawMessage err := json.Unmarshal(b, &objMap if err != nil { return err } var rawAddressMap map[string]*json.RawMessage err = json.Unmarshal(*objMap["address_map"], &rawAddressMap) if err != nil { return err } // how do we unmarshal everything else in the struct, and only override the handling of address map??? // <failing code block is here - beg - just a tad, just a tad, just a tad - recursive> err = json.Unmarshal(b, self) if err != nil { return err } // <failing code block is here - end> if nil == self.AddressMap { self.AddressMap = make(map[string]AddressIf) } for key, value := range rawAddressMap { switch key { case "home": dst := &AddressHome{} err := json.Unmarshal(*value, dst) if err != nil { return err } else { self.AddressMap[key] = dst } } }}我只有name這個 json 示例位中的參數(shù)，但假設(shè)我的代碼中有更多參數(shù)。address_dict有沒有辦法對所有參數(shù)使用普通/默認解組，然后只手動接管?我嘗試了以下方法，并很快意識到為什么它不起作用。 err = json.Unmarshal(b, self) if err != nil { return err }
2 回答
偶然的你
TA貢獻1841條經(jīng)驗 獲得超3個贊
另一個答案的替代方法是聲明一個命名map[string]AddressIf類型并讓它實現(xiàn)json.Unmarshaler接口，然后你不必做任何臨時/匿名類型和轉(zhuǎn)換舞蹈。
type AddressMap map[string]AddressIf
func (m *AddressMap) UnmarshalJSON(b []byte) error {
if *m == nil {
*m = make(AddressMap)
}
raw := make(map[string]json.RawMessage)
if err := json.Unmarshal(b, &raw); err != nil {
return err
}
for key, val := range raw {
switch key {
case "home":
dst := new(AddressHome)
if err := json.Unmarshal(val, dst); err != nil {
return err
}
(*m)[key] = dst
case "work":
dst := new(AddressWork)
if err := json.Unmarshal(val, dst); err != nil {
return err
}
(*m)[key] = dst
}
}
return nil
}
https://play.golang.org/p/o7zV8zu9g5X
收到一只叮咚
TA貢獻1821條經(jīng)驗 獲得超5個贊
為避免復(fù)制聯(lián)系人字段，請使用對象嵌入來隱藏需要特殊處理的字段。
使用工廠模式來消除跨地址類型的代碼重復(fù)。
有關(guān)更多信息，請參閱評論：
var addressFactories = map[string]func() AddressIf{
"home": func() AddressIf { return &AddressHome{} },
"work": func() AddressIf { return &AddressWork{} },
}
func (self *Contact) UnmarshalJSON(b []byte) error {
// Declare type with same base type as Contact. This
// avoids recursion below because this type does not
// have Contact's UnmarshalJSON method.
type contactNoMethods Contact
// Unmarshal to this value. The Contact.AddressMap
// field is shadowed so we can handle unmarshal of
// each value in the map.
var data = struct {
*contactNoMethods
AddressMap map[string]json.RawMessage `json:"address_map"`
}{
// Note that all fields except AddressMap are unmarshaled
// directly to Contact.
(*contactNoMethods)(self),
nil,
}
if err := json.Unmarshal(b, &data); err != nil {
return err
}
// Unmarshal each addresss...
self.AddressMap = make(map[string]AddressIf, len(data.AddressMap))
for k, raw := range data.AddressMap {
factory := addressFactories[k]
if factory == nil {
return fmt.Errorf("unknown key %s", k)
}
address := factory()
if err := json.Unmarshal(raw, address); err != nil {
return err
}
self.AddressMap[k] = address
}
return nil
}
在對該問題的評論中，OP 表示不應(yīng)使用其他類型。這個答案使用了兩種額外contactNoMethods的類型（和匿名類型data）。
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